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Maksim231197 [3]
3 years ago
12

What is the volume of this solid? A. 1104 B. 132 C. 96 D. 276

Mathematics
1 answer:
statuscvo [17]3 years ago
6 0

For this case we have that the volume of the figure is composed of the volume of a prism and the volume of a pyramid:

The volume of the prism is given by:

V = A_ {b} * h

Where:

A_ {b}: It is the area of the base

h: It's the height

Substituting:V = 6 * 6 * 6\\V = 216 \ units ^ 3

The volume of the pyramid is given by:

V = \frac {1} {3} * L ^ 2 * h

Where:

L ^ 2:It is the area of the base

h: It's the height

Substituting:

V = \frac {1} {3} * 6 ^ 2 * 5\\V = \frac {1} {3} * 36 * 5\\V = 60units ^ 3

We add and we have:

V = 276 \ units ^ 3

ANswer:

Option D

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Step-by-step explanation:

Polygons (shapes made of straight lines) where no angle is greater than 180 degrees.  If even one angle is greater than 180 then it is a concave polygon.

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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B36%7D%7B4%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ \frac{36}{4}
iris [78.8K]

Step-by-step explanation:

\sqrt{36/4} =\sqrt{36} /\sqrt{4}

\sqrt{36}=6

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Hope that helps :)

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3 years ago
Divide (- 6 + i)/(1 - 2i) =
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~~~\dfrac{(-6 +i)}{(1- 2i)}\\\\\\=\dfrac{(-6+i)(1+2i)}{(1-2i)(1+2i)}\\\\\\=\dfrac{-6-12i+i+2i^2}{1-(2i)^2}\\\\\\=\dfrac{-6-11i-2}{1+4}~~~~~~~~~~~~~;[i^2=-1]\\\\\\=\dfrac{-8-11i}{5}\\\\\\=-\dfrac 85-\dfrac{11}{5}i

4 0
2 years ago
Read 2 more answers
NEED HELP ASAP!! WILL MARK BRAINLIEST!!!!
dmitriy555 [2]
A = 11/5, let me know if you need me to tell you how i got it (show my work)
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Which of the following relations describes a function?
PtichkaEL [24]

Howdy! My name is Christian and I’ll try and help you with this question!

Date: 9/29/20 Time:  9:08 am CST

Answer:

D.

{ (1, 3), (2, 3), (3, 3), (4, 3) }

Explanation:

This is because for each x value, the number on the left of the set, there is no other x values of the same number, ( 1, 2, 3, 4)

Hope this helps you with your question!

<em>Sincerely, </em>

<em> </em>

<em> </em>

<em>Christian </em>

<em></em>

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